poj2392 Space Elevator（多重背包问题）

Space Elevator

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 8569		Accepted: 4052

Description

The cows are going to space! They plan to achieve orbit by building a sort of space elevator: a giant tower of blocks. They have K (1 <= K <= 400) different types of blocks with which to build the tower. Each block of type i has height h_i (1 <= h_i <= 100) and is available in quantity c_i (1 <= c_i <= 10). Due to possible damage caused by cosmic rays, no part of a block of type i can exceed a maximum altitude a_i (1 <= a_i <= 40000).

Help the cows build the tallest space elevator possible by stacking blocks on top of each other according to the rules.

Input

* Line 1: A single integer, K

* Lines 2..K+1: Each line contains three space-separated integers: h_i, a_i, and c_i. Line i+1 describes block type i.

Output

* Line 1: A single integer H, the maximum height of a tower that can be built

Sample Input

Sample Output

Hint

OUTPUT DETAILS:

From the bottom: 3 blocks of type 2, below 3 of type 1, below 6 of type 3. Stacking 4 blocks of type 2 and 3 of type 1 is not legal, since the top of the last type 1 block would exceed height 40.

题目大意：一群牛要上太空，给出n种石块，每种石块给出单块高度，每种石块都有个限定高度，超过这个高度这种石块就不能被使用了，最后面是每种石块的数量，要求用这些石块能组成的最大高度，并且不能超过限定的高度；

思路：在进行多重背包之前要进行一次排序，将最大高度小的放在前面，只有这样才能得到最优解，如果将大的放在前面，后面有的小的就不能取到，排序之后就可以进行完全背包了

AC代码(一): PZ

 1 #include<stdio.h>
 2 #include<iostream>
 3 #include<algorithm>
 4 #include<string.h>
 5 using namespace std;
 6 #define max 40005
 7 int a[max],b[max],c[max];
 8 struct pp
 9 {
10     int x,y,z;
11 }p[405];
12 int cmp(pp p1,pp p2)//按照限定高度进行排序
13 {
14     return p1.y<p2.y;
15 }
16 int main()
17 {
18     int n;
19     while((scanf("%d",&n))!=EOF)
20     {
21         int i,j,k,t;
22         memset(a,0,sizeof(a));
23         for(i=0;i<n;i++)
24             scanf("%d %d %d",&p[i].x,&p[i].y,&p[i].z);
25         sort(p,p+n,cmp);
26         j=0;
27         //暴力枚举所有的高度，a[s]=1; 表示s这个高度能够达到；
28         for(i=0;i<n;i++)
29         {
30             int s=0,t=1,dd=0;
31             while(s<=p[i].y&&t<=p[i].z) //小于限定的高度，并且小于限定的个数
32             {
33                 s=s+p[i].x;
34                 for(k=0;k<j;k++) 
35                     if(!a[b[k]+s] && (b[k]+s<=p[i].y)) //如果b[k]+s  这个高度没有出现过，并且小于限定的高低，高度可行
36                     {
37                         a[b[k]+s]=1;
38                         c[dd++]=b[k]+s;
39                     }
40                 if(!a[s] && s<=p[i].y) 
41                 {
42                    a[s]=1;
43                    b[j++]=s;
44                 }
45                 t++;              //统计当前石块使用个数
46             }
47             for(k=0;k<dd;k++)
48                 b[j++]=c[k];
49         }
50         int ok=1;
51         for(i=40000;i>=0;i--)
52             if(a[i])
53             {
54                 printf("%d\n",i);
55                 ok=0;
56                 break;
57             }
58         if(ok)
59             printf("0\n");
60     }
61     return 0;
62 }

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